Getting a current going in a circuit

aaaa202 · Sep 4, 2012

I have trouble understanding what happens when you try to get a current going in a circuit. We know that a back emf will be induced when we try to change the current. Therefore the total emf driving current through the wire is given by:
ε₀ - L dI/dt = RI
The solution to this differential equation will be an exponential function going asymptotically towards the final current ε₀/R. I have some trouble understanding the physical reason for this asymptotic behaviour. Intuitively it tells me that in the beginning the current rises a lot due to no backemf, but then the backemf grows and grows and grows until in the end where it is almost equal to that produced by the battery. Is this correct and come someone explain, in some detail, what kind of behaviour this differential equation describes? For instance, I can't see why the back emf should get bigger and bigger.

mfb · Sep 4, 2012

If you consider a resistor and a coil in series:
Low current => low voltage drop at resistor => high voltage drop at coil => quick current change
High current (later) => high voltage drop at resistor => low voltage drop at coil => slow current change

Integral · Sep 4, 2012

Even though it is not clear in the OP you seem to be talking about current though an inductor.

Initially at zero current there is no field surrounding the inductor. When current begins to flow, a field is created by the changing current flow, this field requires energy to build. That field building energy is seen as a resistance to current flow as long as the current is increasing. So the current increases slower then in a purely resistive circuit with similar resistance due to the energy required to build a field around the inductor.

aaaa202 · Sep 4, 2012

Okay, I have understood something completely wrong I think. Let me lay out my understanding of it all:
We have a circuit with a battery that maintains a constant potential difference between the poles. When current starts flowing it will initially be very slow and only a small amount of energt will be lost for the electrons in going through the resistor. That translates to them having a high kinetic + electric potential energy after leaving the resistor. It then sounds from you guys, like this extra energy must be accounted for in a potential difference over the inductor. My question is: WHY IS THIS. Why can't the electrons just pass through the wire and have some extra kinetic energy when they once again arrive at the battery - at the + pole.

mfb · Sep 4, 2012

Do not use kinetic energy for electrons in circuits - it will not lead to anything meaningful. In a very classical picture, electrons are bouncing around all the time at high speeds (>>1000m/s) even without current. A current induces a small bias towards one side - something like millimeters per second. The corresponding momentum is so small that you can neglect it.

The usual assumption is that changes are slow compared to the time light needs to travel within the circuit (if that assumption is wrong, it gets more complicated). In this case, you can always assign a potential to every point of your circuit, and compute voltage drops and so on. The voltage drop in the resistor and the coil together have to add up to the voltage of the battery.

and only a small amount of energt will be lost for the electrons in going through the resistor.

Right.

aaaa202 · Sep 4, 2012

Right. So I think this helped a little but you might want to make it more pictorial.
Let us say we start with a charged battery and hook it up to our circuit. A small current will pass through the resistor and we will have a small drop in potential. The remaining drop in potential must be accounted for in a potential drop over the inductor - is that how I should understand it? Why is that, why can't the electrons not just choose not to deliver energy to the inductor? Try to explain what happens microscopically as a current increases, starting from the very first electrons leaving the minus pole - and don't spare any details :)
Edit: If you know where I can find a simulation that would be great too.

CWatters · Sep 5, 2012

It's not correct to consider current as electrons setting off from the battery. Although electrons are moving about at random rapidly they actually drift along wires very slowly...

The worked example here concludes they move at just 0.00028 meters per second...

http://en.wikipedia.org/wiki/Drift_velocity

Perhaps better to think of it as a water pipe that's already full of water. When you start the pump the pressure increase travels around the pipe faster than the water flows.

Let us say we start with a charged battery and hook it up to our circuit. A small current will pass through the resistor and we will have a small drop in potential. The remaining drop in potential must be accounted for in a potential drop over the inductor - is that how I should understand it?

Correct. The voltages around a closed loop sum to zero. Perhaps better to say the voltage across the inductor starts off large so the remaining small voltage drop appears across the resistor.

Perhaps look at..

http://en.wikipedia.org/wiki/Inductor

When the current flowing through an inductor changes, creating a time-varying magnetic field inside the coil, a voltage is induced, according to Faraday's law of electromagnetic induction, which by Lenz's law opposes the change in current that created it.

aaaa202 · Sep 5, 2012

So how I understand it so far:

A conductor is a material with a lot of free electrons moving around at thermal velocities in random direction. When we put a potential accros one a current will start flowing. When the current is small only a small portion of energy will be lost across the resistor. On the other hand a great deal must then be lost across the inductor. How does this physically happen? Through thermal collisions too? No, definately not since it must be the back-emf that does work but I am unsure how. Please picture it for me! :)

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Getting a current going in a circuit

aaaa202

mfb

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Integral

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mfb

Staff: Mentor

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CWatters

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